The ground state energy of hydrogen atom is – 13 . 6 eV . If the electron jumps to the ground state from the 3 rd excited state, the wavelength of the emitted photon is

Step-by-step solution

The energy of an electron in n th orbit is given by E n = − 13.6 n 2     eV The required energy to jump electron to the ground state from the 3 rd excited state E = E 4 − E 1 = − 13.6 ( 4 ) 2 − [ − 13.6 ( 1 ) 2 ] = − 0.85 + 13.6 = 12.75  eV ∴ The wavelength of the photon emitted is λ = h c E [ ∵ E = h c λ ] = 6.626 × 10 − 34 × 3 × 10 8 12.75 × 1.6 × 10 − 19 = 19.878 × 10 − 7 20.4 = 0.974 ×